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If I take ##<1>## I will get all positive integers. If I take ##<-1>## I will get all negative integers. I should have one element which generates the whole group. What element is this?

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In summary, the conversation discusses the concept of a cyclic group, specifically in the context of the group of integers under addition. The cyclic subgroup generated by an element is defined as the set of all powers of that element, including negative powers in the case of an additive group. Therefore, both 1 and -1 are generators of the group of integers. It is also mentioned that in an infinite cyclic group, an element and its inverse have the same order and are the only possible generators.

- #1

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If I take ##<1>## I will get all positive integers. If I take ##<-1>## I will get all negative integers. I should have one element which generates the whole group. What element is this?

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- #2

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The positive integers do not form a subgroup, so they cannot be the subgroup generated by the element ##1##. Something is wrong with the definitions you are using.LagrangeEuler said:

If I take ##<1>## I will get all positive integers. If I take ##<-1>## I will get all negative integers. I should have one element which generates the whole group. What element is this?

- #3

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If ##G## is additive, ##ng## is written in place of ##g^n##. So when ##G## is additive ##\langle g\rangle = \{ng : n\in \mathbb{Z}\}##. In the case ##G = \mathbb{Z}##, the cyclic subgroup ##\langle 1\rangle = \{n\cdot 1 : n\in \mathbb{Z}\} = \{n : n\in \mathbb{Z}\} = \mathbb{Z}##. Similarly ##\langle -1\rangle = \mathbb{Z}##. So indeed, ##\mathbb{Z}## is cyclic, and both ##-1## and ##1## are generators of the group.

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I think you meanEuge said:If ##G## is abelian,

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Corrected, thanks!PeroK said:I think you meanadditive.

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If a

But if an

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A group element has the same order as its inverse, so for any group ##G## and ##g\in G##, the cyclic subgroup ##\langle g\rangle = \langle g^{-1}\rangle##.Steve4Physics said:

If afinitegroup is cyclic, the group can be generated by a single element.

But if aninfinitegroup (such as integers under addition) is cyclic, the group can be generated by a single elementor its inverse. So in the present case we can use 1 or -1.

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